使用正则表达式 Python 在文本中查找多个字符串

Question

我有以下字符串：

background:url('http://images.bloomingdales.com/is/image/BLM/?&$b=BLM/swatches/&layer=0&size=322,23&src=is{$b$1/optimized/8757901_fpx.tif}&cropN=0,0,14,1&anchor=0,0&layer=1&size=23,23&src=is{$b$2/optimized/8757902_fpx.tif}&anchor=0,0&posN=0.071,0&layer=2&size=23,23&src=is{$b$4/optimized/8234544_fpx.tif}&anchor=0,0&posN=0.143,0&layer=3&size=23,23&src=is{$b$7/optimized/1111977_fpx.tif}&anchor=0,0&posN=0.214,0&layer=4&size=23,23&src=is{$b$0/optimized/8538460_fpx.tif}&anchor=0,0&posN=0.286,0&layer=5&size=23,23&src=is{$b$5/optimized/8234545_fpx.tif}&anchor=0,0&posN=0.357,0&layer=6&size=23,23&src=is{$b$3/optimized/1111973_fpx.tif}&anchor=0,0&posN=0.429,0&layer=7&size=23,23&src=is{$b$7/optimized/1252857_fpx.tif}&anchor=0,0&posN=0.5,0&layer=8&size=23,23&src=is{$b$8/optimized/1252858_fpx.tif}&anchor=0,0&posN=0.571,0&layer=9&size=23,23&src=is{$b$7/optimized/8234547_fpx.tif}&anchor=0,0&posN=0.643,0&layer=10&size=23,23&src=is{$b$0/optimized/8757900_fpx.tif}&anchor=0,0&posN=0.714,0&layer=11&size=23,23&src=is{$b$0/optimized/1111970_fpx.tif}&anchor=0,0&posN=0.786,0&layer=12&size=23,23&src=is{$b$1/optimized/1111971_fpx.tif}&anchor=0,0&posN=0.857,0&layer=13&size=23,23&src=is{$b$2/optimized/1111972_fpx.tif}&anchor=0,0&posN=0.929,0&layer=14&op_sharpen=1&fmt=jpeg&qlt=90,0&hei=23') 322px 0 transparent;

我需要得到所有这些部分：

1/optimized/8757901_fpx.tif、2/optimized/8757902_fpx.tif等。

我正在使用这个正则表达式：

re.findall(re.compile(r'\d{1,2}/optimized/.+\.tif'), swatch)

返回错误结果：

['1/optimized/8757901_fpx.tif}&cropN=0,0,14,1&anchor=0,0&layer=1&size=23,23&src=is{$b$2/optimized/8757902_fpx.tif}&anchor=0,0&posN=0.071,0&layer=2&size=23,23&src=is{$b$4/optimized/8234544_fpx.tif}&anchor=0,0&posN=0.143,0&layer=3&size=23,23&src=is{$b$7/optimized/1111977_fpx.tif}&anchor=0,0&posN=0.214,0&layer=4&size=23,23&src=is{$b$0/optimized/8538460_fpx.tif}&anchor=0,0&posN=0.286,0&layer=5&size=23,23&src=is{$b$5/optimized/8234545_fpx.tif}&anchor=0,0&posN=0.357,0&layer=6&size=23,23&src=is{$b$3/optimized/1111973_fpx.tif}&anchor=0,0&posN=0.429,0&layer=7&size=23,23&src=is{$b$7/optimized/1252857_fpx.tif}&anchor=0,0&posN=0.5,0&layer=8&size=23,23&src=is{$b$8/optimized/1252858_fpx.tif}&anchor=0,0&posN=0.571,0&layer=9&size=23,23&src=is{$b$7/optimized/8234547_fpx.tif}&anchor=0,0&posN=0.643,0&layer=10&size=23,23&src=is{$b$0/optimized/8757900_fpx.tif}&anchor=0,0&posN=0.714,0&layer=11&size=23,23&src=is{$b$0/optimized/1111970_fpx.tif}&anchor=0,0&posN=0.786,0&layer=12&size=23,23&src=is{$b$1/optimized/1111971_fpx.tif}&anchor=0,0&posN=0.857,0&layer=13&size=23,23&src=is{$b$2/optimized/1111972_fpx.tif']

我在 regex101.com 上测试了这个正则表达式，它运行良好： https://regex101.com/r/tV9kU8/1#

Answer 1

re.findall(r'\d{1,2}/optimized/.+?\.tif', swatch)

                                            ^^

通过将? 附加到您的quanitifer 使其不贪婪。

Answer 2

不要使用贪婪的.+，而是在非贪婪模式下使用量词：.+?。 这样，您的正则表达式将永远不会匹配 / 和 .tif 之间超出需要的字符，即它只会匹配到 .tif 的下一个实例。

Answer 3

您可以在您的正则表达式中使用none greedy grouping（请注意，在您的模式中，您还需要在+ 之后放置一个? 以使其成为none greedy）：

>>> re.findall(re.compile(r'{\$b\$(.*?)}'), s)
['1/optimized/8757901_fpx.tif', '2/optimized/8757902_fpx.tif', 
'4/optimized/8234544_fpx.tif', '7/optimized/1111977_fpx.tif', 
'0/optimized/8538460_fpx.tif', '5/optimized/8234545_fpx.tif', 
'3/optimized/1111973_fpx.tif', '7/optimized/1252857_fpx.tif', 
'8/optimized/1252858_fpx.tif', '7/optimized/8234547_fpx.tif', 
'0/optimized/8757900_fpx.tif', '0/optimized/1111970_fpx.tif', 
'1/optimized/1111971_fpx.tif', '2/optimized/1111972_fpx.tif']

由于你们所有人的图像路径都在\$b\$ 之后，您可以使用以下模式：

{\$b\$(.*?)}

这将匹配 {} 内 \$b\$ 之后的任何内容。